INDUCTION MOTORS: ROTATING FIELD, SLIP and TORQUE (part 2)

TORQUE PRODUCTION

In this section we begin with a brief description of rotor types, and introduce the notion of ‘ slip ‘ ,before moving onto explore how the torque is produced, and investigate the variation of torque with speed.

We will find that the behavior of the rotor varies widely according to the slip, and we therefore look separately at low and high values of slip.

Throughout this section we will assume that the rotating magnetic field is unaffected by anything which happens on the rotor side of the air-gap.

Later, we will see that this assumption is pretty well justified.

Rotor construction

Two types of rotor are used in induction motors. In both the rotor ‘iron ‘consists of a stack of steel laminations with evenly spaced slots punched around the circumference. As with the stator laminations, the surface is coated with an oxide layer, which acts as an insulator, preventing unwanted axial eddy currents from flow ing in the iron.

The cage rotor is by far the most common: each rotor slot contains a solid conductor bar and all the conductors are physically and electrically joined together at each end of the rotor by conducting ‘end-rings‘ (see Ill. 10).The conductors may be of copper, in which case the end-rings are brazed-on. Or, in small and medium sizes, the rotor conductors and end-rings can be die cast in aluminum.

The term squirrel cage was widely used at one time and the origin should be clear from Ill. 10.The rotor bars and end-rings are reminiscent of the rotating cages used in bygone days to exercise small rodents (or rather to amuse their human captors).

The absence of any means for making direct electrical connection to the rotor underlines the fact that in the induction motor the rotor currents are induced by the air-gap field. It is equally clear that because the rotor cage comprises permanently short-circuited conductor bars, no external control can be exercised over the resistance of the rotor circuit once the rotor has been made. This is a significant drawback that can be avoided in the second type of rotor, which is known as the ‘wound rotor ‘or ‘slipring ‘type.

In the wound rotor, the slots accommodate a set of three phase windings very much like those on the stator. The windings are connected in star, with the three ends brought out to three sliprings (see Ill. 11).The rotor circuit's thus open, and connection can be made via brushes bearing on the slip-rings. In particular, the resistance of each phase of the rotor circuit can be increased by adding external resistances as indicated in Ill. 11.Adding resistance in appropriate circumstances can be beneficial, as we will see.

Cage rotors are usually cheaper to manufacture, and are very robust and reliable. Until the advent of variable-frequency inverter supplies, however, the superior control which was possible from the slipring type meant that the extra expense of the wound rotor and its associated control gear were frequently justified, especially for high-power machines. Nowadays comparatively few are made, and then only in large sizes. But many old motors remain in service, so they are included in Section 6.

Ill. 10 Cage rotor construction. The stack of pre-punched laminations is shown on the left, with the copper or aluminum rotor bars and end-rings on the right.

Slip

A little thought will show that the behavior of the rotor depends very much on its relative velocity with respect to the rotating field. If the rotor is stationary, for example, the rotating field will cut the rotor conductors at synchronous speed, thereby inducing a high e.m.f. in them. On the other hand, if the rotor was running at the synchronous speed, its relative velocity with respect to the field would be zero, and no e.m.f.'s would be induced in the rotor conductors.

The relative velocity between the rotor and the field is known as the slip. If the speed of the rotor is N, the slip speed is N s N, where N s is the synchronous speed of the field, usually expressed in rev/min. The slip (as distinct from slip speed) is the normalized quantity de fined by

s = [N s - N] / N s [6]

and is usually expressed either as a ratio as in equation 6, or as a percentage. A slip of 0 therefore indicates that the rotor speed is equal to the synchronous speed, while a slip of 1 corresponds to zero speed.

(When tests are performed on induction motors with their rotor deliberately held stationary so that the slip is 1, the test is said to be under 'locked-rotor ‘conditions. The same expression is often used loosely to mean zero speed, even when the rotor is free to move, e.g. when it's started from rest.)

Ill. 11 Schematic diagram of wound rotor for induction motor, showing sliprings and brushes to provide connection to the external (stationary) 3-phase resistance

Rotor induced e.m.f., current and torque

The rate at which the rotor conductors are cut by the flux -and hence their induced e.m.f.-is directly proportional to the slip, with no induced e.m.f. at synchronous speed ( s = 0) and maximum induced e.m.f. when the rotor is stationary ( s = 1).

The frequency of rotor e.m.f. is also directly proportional to slip, since the rotor effectively slides with respect to the flux wave, and the higher the relative speed, the more times in a second each rotor conductor is cut by a N and a S pole. At synchronous speed (slip = 0) the frequency is zero, while at standstill (slip=1), the rotor frequency is equal to the supply frequency. These relationships are shown in Ill. 12.

Although the e.m.f. induced in every rotor bar will have the same magnitude and frequency, they won't be in phase. At any particular instant, bars under the peak of the N poles of the field will have maximum positive voltage in them, those under the peak of the S poles will have maximum negative voltage (i.e.180 dgr phase shift),and those in between will have varying degrees of phase shift. The pattern of instantaneous voltages in the rotor is thus a replica of the flux density wave, and the rotor induced ‘voltage wave‘ therefore moves relative to the rotor at slip speed, as shown in Ill. 13.

Ill. 12 Variation of rotor induced e.m.f and frequency with speed and slip

Ill. 13 Pattern of induced e.m.f.'s in rotor conductors. The rotor ‘ voltage wave ‘ moves at a speed of s N with respect to the rotor surface Since all the rotor bars are short-circuited by the end-rings, the induced voltages will drive currents along the rotor bars, the currents forming closed paths through the end-rings, as shown in the developed diagram (see Ill. 14).

In Ill. 14 the variation of instantaneous e.m.f. in the rotor bars is shown in the upper sketch, while the corresponding instantaneous cur rents flow ing in the rotor bars and end-rings are shown in the lower sketch. The lines representing the currents in the rotor bars have been drawn so that their width is proportional to the instantaneous currents in the bars.

The axial currents in the rotor bars will interact with the radial flux wave to produce the driving torque of the motor, which will act in the same direction as the rotating field, the rotor being dragged along by the field. We note that slip is essential to this mechanism, so that it's never possible for the rotor to catch up with the field, as there would then be no rotor e.m.f., no current and no torque. Finally, we can see that the cage rotor will automatically adapt to whatever pole number is impressed by the stator winding, so that the same rotor can be used for a range of different stator pole numbers.

Rotor currents and torque -small slip

When the slip is small (say between 0 and 10%), the frequency of induced e.m.f. is also very low (between 0 and 5 Hz if the supply frequency is 50 Hz).At these low frequencies the impedance of the rotor circuits is predominantly resistive, the inductive reactance being small because the rotor frequency is low.

Ill. 14 Instantaneous sinusoidal pattern of rotor currents in rotor bars and end-rings. Only one pole-pitch is shown, but the pattern is repeated.

The current in each rotor conductor is therefore in time phase with the e.m.f. in that conductor, and the rotor current wave is therefore in space phase with the rotor e.m.f. wave, which in turn is in space phase with the flux wave. This situation was assumed in the previous discussion, and is represented by the space waveforms shown in Ill. 15.

To calculate the torque we first need to evaluate the ‘B /l r‘ product (see equation (1.2)) to obtain the tangential force on each rotor conductor.

The torque is then given by the total force multiplied by the rotor radius.

We can see from Ill. 15 that where the flux density has a positive peak, so does the rotor current, so that particular bar will contribute a high tangential force to the total torque. Similarly, where the flux has its maximum negative peak, the induced current is maximum and negative, so the tangential force is again positive. We don‘t need to work out the torque in detail, but it should be clear that the resultant will be given by an equation of the form

T = k B / r

where B and I r denote the amplitudes of the flux density wave and the rotor current wave, respectively. Provided that there are a large number of rotor bars (which is a safe bet in practice), the waves shown in Ill. 15 will remain the same at all instants of time, so the torque remains constant as the rotor rotates.

Ill. 15 Pattern of air-gap flux density, induced e.m.f. and current in cage rotor bars at low values of slip If the supply voltage and frequency are constant, the flux will be constant (see equation (5)).The rotor e.m.f. (and hence I r) is then proportional to slip, so we can see from equation (7) that the torque is directly proportional to slip. We must remember that this discussion relates to low values of slip only, but since this is the normal running condition, it's extremely important.

The torque -speed (and torque/slip) relationship for small slips is thus approximately a straight-line, as shown by the section of line AB in Ill. 16.

If the motor is unloaded, it will need very little torque to keep running - only enough to overcome friction in fact -so an unloaded motor will run with a very small slip at just below the synchronous speed, as shown at A in Ill. 16.

When the load is increased, the rotor slows down, and the slip increases, thereby inducing more rotor e.m.f. and current, and thus more torque.

The speed will settle when the slip has increased to the point where the developed torque equals the load torque -- e.g. point B in Ill. 16.

Induction motors are usually designed so that their full-load torque is developed for small values of slip. Small ones typically have a full-load slip of 8%,large ones around 1%.At the full-load slip, the rotor conduct ors will be carrying their safe maximum continuous current, and if the slip is any higher, the rotor will begin to overheat. This overload region is shown by the dotted line in Ill. 16.

The torque -slip (or torque -speed) characteristic shown in Ill. 16 is a good one for most applications, because the speed only falls a little when the load is raised from zero to its full value. We note that, in this normal operating region, the torque -speed curve is very similar to that of a DC motor, which explains why both d.c. and induction motors are often in contention for constant-speed applications.

Rotor currents and torque --large slip

Ill. 16 Torque -speed relationship for low values of slip.

As the slip increases, the rotor e.m.f. and rotor frequency both increase in direct proportion to the slip. At the same time the rotor inductive reactance, which was negligible at low slip (low rotor frequency) begins to be appreciable in comparison with the rotor resistance. Hence, although the induced current continues to increase with slip, it does so more slowly than at low values of slip, as shown in Ill. 17.

At high values of slip, the rotor current also lags behind the rotor e.m.f. because of the inductive reactance. The alternating current in each bar reaches its peak well after the induced voltage, and this in turn means that the rotor current wave has a space-lag with respect to the rotor e.m.f. wave (which is in space phase with the flux wave).This space-lag is shown by the angle f r in Ill. 18.

Ill. 17 Magnitude of current induced in rotor over the full range of slip

Ill. 18 Pattern of air-gap flux density, induced e.m.f. and current in cage rotor bars at high values of slip. (These waveforms should be compared with the corresponding ones when the slip is small, see Ill. 15.) The space-lag means that the peak radial flux density and peak rotor currents no longer coincide, which is bad news from the point of view of torque production, because although we have high values of both flux density and current, they don't occur simultaneously at any point around the periphery. What is worse is that at some points we even have flux density and currents of opposite sign, so over those regions of the rotor surface the torque contributed will actually be negative. The overall torque will still be positive, but is much less than it would be if the flux and current waves were in phase. We can allow for the unwelcome space-lag by modifying equation7, to obtain a more general expression for torque as

T=kBI r cos f r [8]

Equation 7 is merely a special case of equation 8, which only applies under low-slip conditions where cos f r 1.

For most cage rotors, it turns out that as the slip increases the term cos f r reduces more quickly than the current ( I r) increases, so that at some slip between 0 and 1 the developed torque reaches a maximum value. This is illustrated in the typical torque -speed characteristic shown in Ill. 19.The peak torque actually occurs at a slip at which the rotor inductive reactance is equal to the rotor resistance, so the motor designer can position the peak torque at any slip by varying the reactance to resistance ratio.

We will return to the torque-speed curve after we have checked that when we allow for the interaction of the rotor with the stator, our interim conclusions regarding torque production remain valid.

INFLUENCE OF ROTOR CURRENT ON FLUX

Up to now all our discussion has been based on the assumption that the rotating magnetic field remains constant, regardless of what happens on the rotor. We have seen how torque is developed, and that mechanical output power is produced. We have focused attention on the rotor, but the output power must be provided from the stator winding, so we must turn attention to the behavior of the whole motor, rather than just the rotor. Several questions spring to mind.

Ill. 19 Typical complete torque -speed characteristic for cage induction motor Firstly, what happens to the rotating magnetic field when the motor is working? Won‘t the MMF of the rotor currents cause it to change? Secondly, how does the stator know when to start supplying real power across the air-gap to allow the rotor to do useful mechanical work? and f i nally, how will the currents drawn by the stator vary as the slip is changed? These are demanding questions, for which full treatment is beyond our scope. But we can deal with the essence of the matter without too much difficulty. Further illumination can be obtained from study of the equivalent circuit, and this is dealt with in Section 7.

Reduction of flux by rotor current

We should begin by recalling that we have already noted that when the rotor currents are negligible ( s= 0), the e.m.f. that the rotating field induces in the stator winding is very nearly equal to the applied voltage.

Under these conditions a reactive current (which we termed the magnetizing current) flow s into the windings, to set up the rotating flux. Any slight tendency for the flux to fall is immediately detected by a corresponding slight reduction in e.m.f., which is reflected in a disproportionately large increase in magnetizing current, which thus opposes the tendency for the flux to fall.

Exactly the same feedback mechanism comes into play when the slip increases from zero, and rotor currents are induced. The rotor currents are at slip frequency, and they give rise to a rotor MMF wave, which therefore rotates at slip speed ( sN s)relative to the rotor. But the rotor is rotating at a speed of (1 s ) N s, so that when viewed from the stator, the rotor MMF wave always rotates at synchronous speed, regardless of the speed of the rotor.

The rotor MMF wave would, if unchecked, cause its own ‘rotor flux wave‘ ,rotating at synchronous speed in the air-gap, in much the same way that the stator magnetizing current originally set up the flux wave.

The rotor flux wave would oppose the original flux wave, causing the resultant flux wave to reduce.

However, as soon as the resultant flux begins to fall, the stator e.m.f. reduces, thereby admitting more current to the stator winding, and increasing its MMF.A very small drop in the e.m.f. induced in the stator is sufficient to cause a large increase in the current drawn from the mains because the e.m.f. E (see Ill. 8) and the supply voltage V are both very large in comparison with the stator resistance volt drop, IR .The ‘extra‘ stator MMF produced by the large increase in stator current effectively ‘ cancels ‘ the MMF produced by the rotor currents, leaving the resultant MMF (and hence the rotating flux wave)virtually unchanged.

There must be a small drop in the resultant MMF (and flux) of course, to alert the stator to the presence of rotor currents. But because of the delicate balance between the applied voltage and the induced e.m.f. in the stator the change in flux with load is very small, at least over the normal operating speed range, where the slip is small. In large motors, the drop in flux over the normal operating region is typically less than 1%, rising to perhaps 10%in a small motor.

The discussion above should have answered the question as to how the stator knows when to supply mechanical power across the air-gap.

When a mechanical load is applied to the shaft, the rotor slows down, the slip increases, rotor currents are induced and their MMF results in a modest (but vitally important) reduction in the air-gap flux wave. This in turn causes a reduction in the e.m.f. induced in the stator windings and therefore an increase in the stator current drawn form the supply. We can anticipate that this is a stable process (at least over the normal operating range) and that the speed will settle when the slip has increased sufficiently that the motor torque equals the load torque.

As far as our conclusions regarding torque are concerned, we see that our original assumption that the flux was constant is near enough correct when the slip is small. We will find it helpful and convenient to continue to treat the flux as constant (for given stator voltage and frequency) when we turn later to methods of controlling the normal running speed.

It has to be admitted, however, that at high values of slip (i.e. low rotor speeds), we cannot expect the main flux to remain constant, and in fact we would find in practice that when the motor was first switched-on, with the rotor stationary, the main flux might typically be only half what it was when the motor was at full speed. This is because at high slips, the leakage fluxes assume a much greater importance than under normal low-slip conditions. The simple arguments we have advanced to predict torque would therefore need to be modified to take account of the reduction of main flux if we wanted to use them quantitatively at high slips. There is no need for us to do this explicitly, but it will be reflected in any subsequent curves portraying typical torque -speed curves for real motors. Such curves are of course used when selecting a motor, since they provide the easiest means of checking whether the starting and run up torque is adequate for the job in hand.

STATOR CURRENT-SPEED CHARACTERISTICS

In the previous section, we argued that as the slip increased, and the rotor did more mechanical work, the stator current increased. Since the extra current is associated with the supply of real (i.e. mechanical output)power (as distinct from the original magnetizing current which was seen to be reactive),this additional ‘ work ‘ component of current is more or less in phase with the supply voltage, as shown in the phasor diagrams (Ill. 20).

The resultant stator current is the sum of the magnetizing current, which is present all the time, and the load component, which increases with the slip. We can see that as the load increases, the resultant stator current also increases, and moves more nearly into phase with the voltage. But because the magnetizing current is appreciable, the difference in magnitude between no-load and full-load currents may not be all that great. (This is in sharp contrast to the DC motor, where the no-load current in the armature is very small in comparison with the full load current. Note, however, that in the DC motor, the excitation ( flux) is provided by a separate field circuit, whereas in the induction motor the stator winding furnishes both the excitation and the work currents. If we consider the behavior of the work components of current only, both types of machine look very similar.) The simple ideas behind Ill. 20 are based on an approximation, so we cannot push them too far: they are fairly close to the truth for the normal operating region, but breakdown at higher slips, where the rotor and stator leakage reactances become significant. A typical current locus over the whole range of slips for a cage motor is shown in Ill. 21.

We note that the power factor becomes worse again at high slips, and also that the current at standstill (i.e. the ‘starting‘ current) is perhaps five times the full-load value.

Ill. 20 Phasor diagrams showing stator current at no-load, part-load and full-The resultant current in each case is the sum of the no-load (magnetizing) current and load component Very high starting currents are one of the worst features of the cage induction motor. They not only cause unwelcome volt drops in the supply system, but also call for heavier switchgear than would be needed to cope with full-load conditions. Unfortunately, for reasons discussed earlier, the high starting currents are not accompanied by high starting torques, as we can see from Ill. 22, which shows current and torque as functions of slip for a general-purpose cage motor.

We note that the torque per ampere of current drawn from the mains is typically very low at start up, and only reaches a respectable value in the normal operating region, i.e. when the slip is small. This matter is explored further in Section 6, and also by means of the equivalent circuit in Section 7.

Ill. 21 Phasor diagram showing the locus of stator current over the full range of speeds from no-load (full speed) down to the locked-rotor (starting) condition Ill. 22 Typical torque-speed and current -speed curves for a cage induction motor.

The torque and current axes are scaled so that 100% represents the continuously rated (full-load) value.

QUIZ

The first ten questions are designed to reinforce understanding by means of straightforward calculations, while the remainder are more demanding and involve some extension of the basic ideas.

1) At what frequency must a 4-pole motor be supplied so that its synchronous speed is 1200 rev/min?

2) The name Image of a standard 50 Hz induction motor quotes full load speed as 2950 rev/min. Find the pole number and the rated slip.

3) A 4-pole,60 Hz induction motor runs with a slip of 4%. Find: (a) the speed; (b)the rotor frequency;(c)the speed of rotation of the rotor current wave relative to the rotor surface;(d)the speed of rotation of the rotor current wave relative to the stator surface.

4) Choose a suitable pole number for an induction motor to cover the speed range from 400 rev/min to 800 rev/min when supplied from a 30 -75 Hz variable-frequency source.

5) The r.m.s. current in the rotor bars of an induction motor running with a slip of 1% is 25 A, and the torque produced is 20 N m.

Estimate the rotor current and torque when the load is increased so that the motor slip is 3%.

6) An induction motor designed for operation from 440ffis supplied at 380ffinstead.What effect will the reduced voltage have on the following:(a)the synchronous speed;(b)the magnitude of the air-gap flux;(c)the induced current in the rotor when running at rated speed;(d)the torque produced at rated speed.

7) The rotor from a 6-pole induction motor is to be used with a 4-pole stator having the same bore and length as the 6-pole stator. What modifications would be required to the rotor if it was (a) a squirrel cage type, and (b) a wound-rotor type?

8) A 440 V,60 Hz induction motor is to be used on a 50 Hz supply. What voltage should be used?

9) The stator of a 220ffinduction motor is to be rewound for operation from a 440 ff supply. The original coils each had 15 turns of 1 mm diameter wire. Estimate the number of turns and diameter of wire for the new stator coils.

10) The slip of an induction motor driving a constant torque load is 2.0%.If the voltage is reduced by 5%, estimate: (a) the new steady-state rotor current, expressed in terms of its original value ;(b) the new steady-state slip.

11) Why can an induction motor never run at its synchronous speed?

12) Why is it important to maintain the ratio of voltage: frequency at the correct value for an induction motor? What would be the consequences of making the ratio (a) higher, and (b) lower, than the specified value?

13)An induction motor with a 2 mm air-gap has a no-load (magnetizing)current of 5 A. Assuming that the air-gap represents the only significant reluctance in the path of the main flux, explain why, if the rotor was reground so that the air-gap became 3 mm, the magnetizing current would be expected to increase to 7.5 A.

What effect would the increase in air-gap be expected to have on the magnitude of the air-gap flux wave?

14) As the slip of an induction motor increases, the current in the rotor increases, but beyond a certain slip the torque begins to fall. Why is this?

15) For a given rotor diameter, the stator diameter of a 2-pole motor is much greater than the stator diameter of, say, an 8-pole motor. By sketching and comparing the magnetic flux patterns for machines with low and high pole numbers, explain why more stator iron is required as the pole number reduces.

16) The layout of coils for 4-pole and 6-pole windings in a 36-slot stator are shown in Ill. 6. All the coils have the same number of turns of the same wire, the only real difference being that the 6 pole coils are of shorter pitch.

Sketch the MMF wave produced by one phase, for each pole number, assuming that the same current flow s in every coil. Hence show that to achieve the same amplitude of flux wave, the current in the 6-pole winding would have to be about 50%larger than in the 4-pole.

How does this exercise relate to the claim that the power factor of a low-speed induction motor is usually lower than a high-speed one?